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根号(a^2+b^2)+根号(b^2+c^2)>=根号2(a+b+c)
人气:439 ℃ 时间:2020-01-30 05:26:36
解答
右边变形为:根号2(a+b)/2 + 根号2(b+c)/2+根号2(a+c)/2然后利用基本不等式得:根号2(a+b)/2 >=根号a^2+b^2 (算数平均数小于平方平均数)同理:根号2(a+c)/2 >=根号a^2+c^2根号2(b+c)/2 >=根号b^2+c^2原题得证...
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